### Multinomial Goodness of Fit Hypothesis Test **Scenario:** You are conducting a multinomial Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. That is: \[ H_0 : p_A = p_B = p_C = p_D = p_E \] ### Table Completion: Complete the table. Report all answers correct to three decimal places. | Category | Observed Count | Expected Count | Component Value | |----------|----------------|----------------|-----------------| | A | 7 | | | | B | 21 | | | | C | 19 | | | | D | 9 | | | | E | 15 | | | ### Calculations: 1. **Expected Count:** Given that the null hypothesis states that all categories are equally likely, you need to find the expected count for each category. \[ \text{Expected Count} = \frac{\text{Total Observed Count}}{\text{Number of Categories}} \] 2. **Component Value:** For each category, calculate the component value as follows: \[ \text{Component Value} = \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}} \] 3. **Total Chi-Square Test-Statistic:** Sum all the component values to get the chi-square test statistic \(\chi^2\). \[ \chi^2 = \sum \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}} \] ### Hypothesis Test Results: - **Chi-square test-statistic (\(\chi^2\)):** \[ \chi^2 = \_\_\_\_\_ \] - **p-value:** \[ \text{p-value} = \_\_\_\_\_ \] ### Conclusion: For significance level \(\alpha = 0.005\), what would be the conclusion of this hypothesis test? - [ ] Reject the Null Hypothesis - [ ] Fail to reject the Null Hypothesis **Note:** - The chi-square distribution table or software can be used to find the p-value corresponding to the calculated \
### Multinomial Goodness of Fit Hypothesis Test **Scenario:** You are conducting a multinomial Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. That is: \[ H_0 : p_A = p_B = p_C = p_D = p_E \] ### Table Completion: Complete the table. Report all answers correct to three decimal places. | Category | Observed Count | Expected Count | Component Value | |----------|----------------|----------------|-----------------| | A | 7 | | | | B | 21 | | | | C | 19 | | | | D | 9 | | | | E | 15 | | | ### Calculations: 1. **Expected Count:** Given that the null hypothesis states that all categories are equally likely, you need to find the expected count for each category. \[ \text{Expected Count} = \frac{\text{Total Observed Count}}{\text{Number of Categories}} \] 2. **Component Value:** For each category, calculate the component value as follows: \[ \text{Component Value} = \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}} \] 3. **Total Chi-Square Test-Statistic:** Sum all the component values to get the chi-square test statistic \(\chi^2\). \[ \chi^2 = \sum \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}} \] ### Hypothesis Test Results: - **Chi-square test-statistic (\(\chi^2\)):** \[ \chi^2 = \_\_\_\_\_ \] - **p-value:** \[ \text{p-value} = \_\_\_\_\_ \] ### Conclusion: For significance level \(\alpha = 0.005\), what would be the conclusion of this hypothesis test? - [ ] Reject the Null Hypothesis - [ ] Fail to reject the Null Hypothesis **Note:** - The chi-square distribution table or software can be used to find the p-value corresponding to the calculated \
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
You are conducting a multinomial Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. That is:
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Complete the table. Report all answers correct to three decimal places.
| Category | Observed Count |
Expected Count |
Component Value |
|---|---|---|---|
| A | 7 | ||
| B | 21 | ||
| C | 19 | ||
| D | 9 | ||
| E | 15 |
What is the chi-square test-statistic for this data?
�2=
What is the �-value?
�-value =
For significance level �=0.005, what would be the conclusion of this hypothesis test?
- Reject the Null Hypothesis
- Fail to reject the Null Hypothesis
Report all answers accurate to three decimal places.
![### Multinomial Goodness of Fit Hypothesis Test
**Scenario:**
You are conducting a multinomial Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. That is:
\[ H_0 : p_A = p_B = p_C = p_D = p_E \]
### Table Completion:
Complete the table. Report all answers correct to three decimal places.
| Category | Observed Count | Expected Count | Component Value |
|----------|----------------|----------------|-----------------|
| A | 7 | | |
| B | 21 | | |
| C | 19 | | |
| D | 9 | | |
| E | 15 | | |
### Calculations:
1. **Expected Count:**
Given that the null hypothesis states that all categories are equally likely, you need to find the expected count for each category.
\[
\text{Expected Count} = \frac{\text{Total Observed Count}}{\text{Number of Categories}}
\]
2. **Component Value:**
For each category, calculate the component value as follows:
\[
\text{Component Value} = \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}}
\]
3. **Total Chi-Square Test-Statistic:**
Sum all the component values to get the chi-square test statistic \(\chi^2\).
\[
\chi^2 = \sum \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}}
\]
### Hypothesis Test Results:
- **Chi-square test-statistic (\(\chi^2\)):**
\[
\chi^2 = \_\_\_\_\_
\]
- **p-value:**
\[
\text{p-value} = \_\_\_\_\_
\]
### Conclusion:
For significance level \(\alpha = 0.005\), what would be the conclusion of this hypothesis test?
- [ ] Reject the Null Hypothesis
- [ ] Fail to reject the Null Hypothesis
**Note:**
- The chi-square distribution table or software can be used to find the p-value corresponding to the calculated \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81fcbcae-c7c7-47f7-b9b9-962fae442be8%2Ffaaeea00-464b-4b1b-b4a8-3dfd2c22d930%2F2w7jeyk_processed.png&w=3840&q=75)
Transcribed Image Text:### Multinomial Goodness of Fit Hypothesis Test
**Scenario:**
You are conducting a multinomial Goodness of Fit hypothesis test for the claim that all 5 categories are equally likely to be selected. That is:
\[ H_0 : p_A = p_B = p_C = p_D = p_E \]
### Table Completion:
Complete the table. Report all answers correct to three decimal places.
| Category | Observed Count | Expected Count | Component Value |
|----------|----------------|----------------|-----------------|
| A | 7 | | |
| B | 21 | | |
| C | 19 | | |
| D | 9 | | |
| E | 15 | | |
### Calculations:
1. **Expected Count:**
Given that the null hypothesis states that all categories are equally likely, you need to find the expected count for each category.
\[
\text{Expected Count} = \frac{\text{Total Observed Count}}{\text{Number of Categories}}
\]
2. **Component Value:**
For each category, calculate the component value as follows:
\[
\text{Component Value} = \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}}
\]
3. **Total Chi-Square Test-Statistic:**
Sum all the component values to get the chi-square test statistic \(\chi^2\).
\[
\chi^2 = \sum \frac{(\text{Observed Count} - \text{Expected Count})^2}{\text{Expected Count}}
\]
### Hypothesis Test Results:
- **Chi-square test-statistic (\(\chi^2\)):**
\[
\chi^2 = \_\_\_\_\_
\]
- **p-value:**
\[
\text{p-value} = \_\_\_\_\_
\]
### Conclusion:
For significance level \(\alpha = 0.005\), what would be the conclusion of this hypothesis test?
- [ ] Reject the Null Hypothesis
- [ ] Fail to reject the Null Hypothesis
**Note:**
- The chi-square distribution table or software can be used to find the p-value corresponding to the calculated \
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